It is well-known that a pair of pants can be given a constant negative curvature metric. Indeed, one way to see that the genus $g$ surface can be given a constant negative curvature metric for $g \geq 2$ is to use the fact that each such surface can be partitioned into pants. Here's a picture of a pants-decomposition of the genus-3 surface. Image credit: this blog post of Elizabeth Denne.
Now, when I draw a pair of pants embedded in Euclidean 3-space, it always looks to me as though there is some point, around the area where the three legs are joined, where the curvature becomes non-negative.
Question: Can we embed a pair of pants in $\mathbb{R}^3$ in such a way that the inherited metric has everywhere-negative curvature?
Added: The formulation of this question still needs some work. At the moment it admits a rather "cheap" solution. One can take, for instance a negatively curved cylinder (easy to construct as a surface of revolution) and then delete a disk from that. I think I want to add some hypothesis along the lines:
Each of the boundary circles of the pants is a geodesic.